#ATTENDANCE QUIZ for Lecture 7 of Math251(Dr. Z.)
#EMAIL  RIGHT AFTER YOU WATCHED THE VIDEO
#BUT NO LATER THAN  Sept. 28, 2020, 8:00PM (Rutgers time)
#THIS .txt FILE (EDITED WITH YOUR ANSWERS)
#TO:
#DrZcalc3@gmail.com
#Subject: aq7
#with an ATTACHMENT CALLED:
#aq7FirstLast.txt
#(e.g. aq7DoronZeilberger.txt)




#LIST ALL THE ATTENDANCE QUESTIONS FOLLOWED BY THEIR ANSWERS


Q1. THE FIRST ATTENDANCE QUESTION WAS:

Let a =5th digit of your RUID
b=2nd digit of your RUID 
Let f(x,y)=x^a+bx^2a*y^3
Find the two partial derivatives of f(x,y)
A1. MY ANSWER TO THE FIRST ATTENDANCE QUESTION IS:
a=0 b=0 
f(x,y)=x^0+0=1 does not exist


Q2. THE  SECOND ATTENDANCE QUESTION WAS:
With a and b as a above find f-x(z(x,y)) and f-y(z(x,y))
If x,y,z are related by the relationship (Equation)
x^a*y^2*z^b+x^2*y^3b*z^3=exp(x*y*z)

A2. MY ANSWER TO THE  SECOND ATTENDANCE QUESTION IS:
a=0 b=0
x^0*y^2*z^0+x^2*y^0*z^3=exp(0)
1*y^2*1+x^2*1*z^3=exp(0)
y^2+x^2*z^3=exp(0)=1
y^2+x^2*z^3=1
z/x=2x*z^3+x^2*3z^2*z'=0
z'=-2x*z^3/x^2*3z^2
z/y=2y+x^2*3z^2*z'
z'=-2y/x^2*3z^2


Q3. THE  THIRD ATTENDANCE QUESTION WAS:
Do the second part, Find f-y(1,1)
x^2+y^2=xyz+xyz^5
A3. MY ANSWER TO THE  THIRD ATTENDANCE QUESTION IS:
2y=y*(xz)'+y*(xz^5)'
2y=y*(z+xz')+y*(z^5+x5z^4z')
f-y(1,1)
2*1=1*(1+z')+1*(1^5+5*1^4z')
2=1+z'+	1^5+5*1^4z'=2+6z'
z'=0

Q4. THE FIRST ATTENDANCE QUESTION WAS:
Is the answer the same, is  f-y(1,1) also 0?

A4. MY ANSWER TO THE FIRST ATTENDANCE QUESTION IS:
Yes, the answer is the same. 
f-y(1,1) is also 0.


Q5. THE  SECOND ATTENDANCE QUESTION WAS:
Let a and b be as above Find the equation of the tangent 
plane to the surface Z=x^a+y^b+a*b*x^y at the point (1,1,2+abŁ©

A5. MY ANSWER TO THE  SECOND ATTENDANCE QUESTION IS:
a=0 b=0 
Z=x^0+y^0+0*0*x^y=1+1+0=2
does not exist.